Question

28. In Problems 1-30 use appropriate algebra and Theorem 7.2.1 to find the given inverse Laplace transform.

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{s^4-9}\right\}$

Short answer

$\frac{1}{6\sqrt{3}}\sinh \left(\sqrt{3}t\right)-\frac{1}{6\sqrt{3}}\sin \left(\sqrt{3}t\right)$

Step-by-step solution

Definition of inverse Transform

THEOREM 7.2.1 Some Inverse Transforms

1. $1={\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}$
   
2. ${\mathrm{t}}^{\mathrm{n}}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{n}!}{{\mathrm{s}}^{\mathrm{n}+1}}\right\},\mathrm{n}=1,2,3,\dots$
   
3. ${\mathrm{e}}^{\mathrm{at}}={\mathcal{L}}^{-1}\left\{\frac{1}{\mathrm{s}-\mathrm{a}}\right\}$
   
4. $\mathrm{sinkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
   
5. $\mathrm{coskt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
   
6. $\mathrm{sinhkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   
7. $\mathrm{coshkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   

Use Inverse Transform

${\mathrm{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^4-9}\right\}={\mathrm{L}}^{-1}\left\{\frac{1}{\left({\mathrm{s}}^2+3\right)\left({\mathrm{s}}^2-3\right)}\right\}$

Here we have to decompose the fraction.

Decomposing

$\frac{1}{\left({\mathrm{s}}^2+3\right)\left({\mathrm{s}}^2+3\right)}=\frac{\mathrm{as}+\mathrm{b}}{{\mathrm{s}}^2+3}+\frac{\mathrm{cs}+\mathrm{d}}{{\mathrm{s}}^2-3}\to \frac{1}{\left({\mathrm{s}}^2+3\right)\left({\mathrm{s}}^2-3\right)}=\frac{1}{6\left({\mathrm{s}}^2-3\right)}-\frac{1}{6\left({\mathrm{s}}^2+3\right)}$

The trick is to algebraically manipulate the expression make it look like something we know

Now,$\frac{1}{6}{\mathrm{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^2-3}\right\}-\frac{1}{6}{\mathrm{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^2+3}\right\}$.

For,$-\frac{1}{6}\frac{1}{\sqrt{3}}\frac{\sqrt{3}}{s^2+3}=-\frac{1}{6}\frac{1}{s^2+3}$

And,$\frac{1}{6}\frac{1}{\sqrt{3}}\frac{\sqrt{3}}{s^2-3}=\frac{1}{6}\frac{1}{s^2-3}$

So,$=\frac{1}{6\sqrt{3}}\sinh \left(\sqrt{3}t\right)-\frac{1}{6\sqrt{3}}\sin \left(\sqrt{3}t\right)$.